[size=150]The Taylor series of a function is a representation of the function as an infinite sum of its derivatives evaluated at a specific point. The Taylor series allows us to approximate a function with a polynomial, which becomes more accurate as we include more terms in the series. The Taylor series of a function (f(x)) centered at (x = a) is given by:[/size][br][br][math]f\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)+\frac{f''\left(a\right)}{2!}\left(x-a\right)^2+\frac{f'''\left(a\right)}{3!}\left(x-a\right)^3+\frac{f''''\left(a\right)}{4!}\left(x-a\right)^4+...[/math][br][br][size=150]In this representation, f(a) is the value of the function at the center x = a, f'(a) is the first derivative of the function evaluated at x = a, f''(a) is the second derivative evaluated at x = a, and so on.[br][br][/size][size=150]Let's find the Taylor series of the secant function (sec(x)) centered at x = 0. First, we need to find the derivatives of (sec(x)):[br][/size][br][math]sec\left(x\right)=\frac{1}{cos\left(x\right)}[/math][br][br][size=150]Taking the derivative of (sec(x)) with respect to (x), we get:[/size][br][br][math]sec'\left(x\right)=\frac{d}{dx}\left(\frac{1}{cos\left(x\right)}\right)=-\frac{sin\left(x\right)}{cos^2\left(x\right)}=-sec\left(x\right)\cdot tan\left(x\right)[/math][br][br][size=150]The second derivative of (sec(x) is:[/size][br][br][math]sec''\left(x\right)=\frac{d}{dx}\left(-sec\left(x\right)\cdot tan\left(x\right)\right)=sec'\left(x\right)\cdot tan\left(x\right)-sec\left(x\right)\cdot sec^2\left(x\right)=-sec\left(x\right)\cdot\left(sec\left(x\right)\cdot tan\left(x\right)+tan\left(x\right)^2\right)[/math][br][br][size=150]Evaluating (sec''(0)):[/size][br][br][math]sec''\left(0\right)=-sec\left(0\right)\cdot\left(sec\left(0\right)\cdot tan\left(0\right)+tan\left(0\right)^2\right)=-1\cdot\left(1\cdot0+0^2\right)=0[/math][br][br][size=150]All the higher-order derivatives of (sec(x)) evaluated at (x = 0) are also 0. Therefore, the Taylor series of (sec(x)) centered at (x = 0) is:[/size][br][br][math]sec\left(x\right)=sec\left(0\right)+sec'\left(0\right)\cdot x+\frac{sec''\left(0\right)}{2!}+x^2+\frac{sec'''\left(0\right)}{3!}+x^3+...[/math][br][br][size=150]Since all the higher-order derivatives are 0, the Taylor series simplifies to:[br][br]sec(x) = sec(0) + sec'(0) * x[/size][br][size=150][br]We can evaluate (sec(0)) and (sec'(0)) to find the final form of the Taylor series:[/size][br][br][math]sec\left(0\right)=\frac{1}{cos\left(0\right)}=1[/math][br][br][math]sec'\left(0\right)=-sec\left(0\right)\cdot tan\left(0\right)=-1\cdot0=0[/math][br][br][size=150]Therefore, the Taylor series of the secant function centered at (x = 0) is:[/size][br][br][math]sec\left(x\right)=1+0\cdot x=1[/math][br][br][size=150]So, the secant function can be approximated by the constant value of 1 when expanded as a Taylor series centered at (x = 0).[/size]